转移矩阵+快速幂HDU5607

题意：http://acm.hdu.edu.cn/showproblem.php?pid=5607

有向图中问你从定点u走到x的概率

思路：

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr strcat
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
#include <iomanip>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
clock_t __START,__END;
double __TOTALTIME;
void _MS(){__START=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const db E=2.718281828;
const db PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const db ESP=1e-9;
const int mod=(int)1e9+7;
const int N=60;

struct Mat
{
    ll mat[N][N];
    Mat operator*(const Mat &a)const
    {
        Mat b=Mat();
        for(int i=1;i<=55;++i)
        {
            for(int j=1;j<=55;++j)
            {
                for(int k=1;k<=55;++k)
                {
                    b.mat[i][j]=(b.mat[i][j]+mat[i][k]*a.mat[k][j]%mod);
                    b.mat[i][j]%=mod;
                }
            }
        }
        return b;
    }
}start;
Mat Mqpow(long long n,int x)
{
    Mat ans=Mat();
    for(int i=1;i<=x;++i)
        ans.mat[i][i]=1;
    while(n)
    {
        if(n&1)
            ans=ans*start;
        start=start*start;
        n>>=1;
    }
    return ans;
}
ll qpow(ll a,ll b,ll mod)
{
    ll ans;
//    a%=mod;
    ans=1;
    while(b!=0)
    {
        if(b&1)
            ans=(a*ans)%mod;//注意左乘右乘;
        b/=2;
        a=(a*a)%mod;
    }
    return ans;
}
ll cnt[N];

int main()
{
    int n,m;
    while(~sc("%d%d",&n,&m))
    {
        mem(cnt,0);
        start=Mat();
        for(int i=1;i<=m;++i)
        {
            int u,v;
            sc("%d%d",&u,&v);
            start.mat[u][v]++;
            cnt[u]++;
        }
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                if(start.mat[i][j])
                    start.mat[i][j]=start.mat[i][j]*qpow(cnt[i],mod-2,mod)%mod;
        int ask;
        sc("%d",&ask);
        Mat temp=start;
        for(int i=1;i<=ask;++i)
        {
            int u,k;
            sc("%d%d",&u,&k);
            start=temp;
            Mat ans=Mqpow(k,n);
            for(int j=1;j<=n;++j)
                pr("%lld ",ans.mat[u][j]);
            pr("
");
        }
    }
    return 0;
}

/**************************************************************************************/